HOMEWORK 06

SECTION 3.2

Assigned: 9, 11, 17

3.2.9

Problem:

Consider the system of differential equations given by

x_1' = -x_1 + 2x_2

x_2' = -7x_1 + 8x_2

1. Determine a matrix \mathbf{A} so that the system may be written in the form \mathbf{x}' = \mathbf{A} \mathbf{x}.

2. Determine all constant solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.

3. Compute the eigenvalues and eigenvectors of \mathbf{A}.

4. Determine all straight-line solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.

5. Find a more general solution to \mathbf{x}' = \mathbf{A} \mathbf{x} by taking all possible linear combinations of the straight-line solutions from (d).

6. Solve the initial-value problem \mathbf{x}' = \mathbf{A} \mathbf{x}, \mathbf{x}(0) = \begin{bmatrix} -2 \\ 0 \end{bmatrix}. Discuss the graphical behavior of this solution.

Solution:

(a) Determine the Matrix \mathbf{A}

The system is:

x_1' = -x_1 + 2x_2

x_2' = -7x_1 + 8x_2

This can be written in matrix form as:

\mathbf{x}' = \begin{bmatrix} -1 & 2 \\ -7 & 8 \end{bmatrix} \mathbf{x}

So, the matrix \mathbf{A} is:

\mathbf{A} = \begin{bmatrix} -1 & 2 \\ -7 & 8 \end{bmatrix}

(b) Determine All Constant Solutions

Constant solutions satisfy \mathbf{x}' = \mathbf{0}, so:

\mathbf{A} \mathbf{x} = \mathbf{0}

This leads to the homogeneous system:

- x_1 + 2 x_2 = 0

-7 x_1 + 8 x_2 = 0

Solving, we find that x_1 = x_2 = 0. So, the only constant solution is:

\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

(c) Compute the Eigenvalues and Eigenvectors of \mathbf{A}

Compute the characteristic equation:

\det(\mathbf{A} - \lambda \mathbf{I}) = 0

\begin{vmatrix} -1 - \lambda & 2 \\ -7 & 8 - \lambda \end{vmatrix} = (-1 - \lambda)(8 - \lambda) - (-7)(2) = 0

Simplify:

\lambda^2 - 7\lambda + 6 = 0 \implies (\lambda - 6)(\lambda - 1) = 0

Eigenvalues are \lambda_1 = 6 and \lambda_2 = 1.

For \lambda = 1:

(\mathbf{A} - \mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -2 & 2 \\ -7 & 7 \end{bmatrix} \mathbf{v} = \mathbf{0}

This yields v_1 = v_2. So, an eigenvector is:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

For \lambda = 6:

(\mathbf{A} - 6\mathbf{I})\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -7 & 2 \\ -7 & 2 \end{bmatrix} \mathbf{v} = \mathbf{0}

This gives v_2 = \frac{7}{2}v_1. An eigenvector is:

\mathbf{v}_2 = \begin{bmatrix} 2 \\ 7 \end{bmatrix}

(d) Determine All Straight-Line Solutions

Straight-line solutions are of the form \mathbf{x}(t) = \mathbf{v} e^{\lambda t}:

\mathbf{x}_1(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{t}, \quad \mathbf{x}_2(t) = c_2 \begin{bmatrix} 2 \\ 7 \end{bmatrix} e^{6t}

(e) General Solution as a Linear Combination of Straight-Line Solutions

\mathbf{x}(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 2 \\ 7 \end{bmatrix} e^{6t}

(f) Solve the Initial-Value Problem

Initial condition \mathbf{x}(0) = \begin{bmatrix} -2 \\ 0 \end{bmatrix}:

\begin{cases} c_1 + 2c_2 = -2 \\ c_1 + 7c_2 = 0 \end{cases}

Solving:

c_2 = \frac{2}{5}, \quad c_1 = -\frac{14}{5}

So the solution is:

\mathbf{x}(t) = -\frac{14}{5} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{t} + \frac{2}{5} \begin{bmatrix} 2 \\ 7 \end{bmatrix} e^{6t}

Graphical Behavior of the Solution

• As t \to \infty, e^{6t} dominates e^{t}, so \mathbf{x}(t) grows exponentially in the direction of \begin{bmatrix} 2 \\ 7 \end{bmatrix}.

Final Answer:

(a) An explicit matrix \mathbf{A} is

\mathbf{A} = \begin{bmatrix} -1 & 2 \\ -7 & 8 \end{bmatrix}

(b) All constant solutions satisfy

\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

(c) Eigenvalues are \lambda = 6 and \lambda = 1; corresponding eigenvectors are

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\quad \mathbf{v}_2 = \begin{bmatrix} 2 \\ 7 \end{bmatrix}

(d) All straight-line solutions are

\mathbf{x}(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 2 \\ 7 \end{bmatrix} e^{6t}

(e) The general solution is the linear combination of straight-line solutions:

\mathbf{x}(t) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{t} + c_2 \begin{bmatrix} 2 \\ 7 \end{bmatrix} e^{6t}

(f) Solving the initial-value problem gives:

\mathbf{x}(t) = -\frac{14}{5} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{t} + \frac{2}{5} \begin{bmatrix} 2 \\ 7 \end{bmatrix} e^{6t}

Graphical behavior: As t \to \infty, e^{6t} dominates e^{t}, so \mathbf{x}(t) grows exponentially in the direction of \begin{bmatrix} 2 \\ 7 \end{bmatrix}.

3.2.11

Problem:

Consider the system of differential equations given by

x_1' = -2x_1 + x_2

x_2' = -2x_2

1. Determine a matrix \mathbf{A} so that the system may be written in the form \mathbf{x}' = \mathbf{A} \mathbf{x}.

2. Determine all constant solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.

3. Compute the eigenvalues and eigenvectors of \mathbf{A}.

4. Determine all straight-line solutions to \mathbf{x}' = \mathbf{A} \mathbf{x}.

5. Find a more general solution to \mathbf{x}' = \mathbf{A} \mathbf{x} by taking all possible linear combinations of the straight-line solutions from (d).

6. Attempt to solve the initial-value problem \mathbf{x}' = \mathbf{A} \mathbf{x}, \mathbf{x}(0) = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. What does this tell you about the proposed general solution in (e)?

Solution:

(a) Determine the Matrix \mathbf{A}

The system is:

x_1' = -2x_1 + x_2

x_2' = -2x_2

This can be written in matrix form as:

\mathbf{x}' = \begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix} \mathbf{x}

So, the matrix \mathbf{A} is:

\mathbf{A} = \begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}

(b) Determine All Constant Solutions

Constant solutions satisfy \mathbf{x}' = \mathbf{0}, so:

\mathbf{A} \mathbf{x} = \mathbf{0}

This leads to the homogeneous system:

-2 x_1 + x_2 = 0

-2 x_2 = 0

From the second equation, x_2 = 0. Substituting into the first equation gives x_1 = 0. Therefore, the only constant solution is:

\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

(c) Compute the Eigenvalues and Eigenvectors of \mathbf{A}

Compute the characteristic equation:

\det(\mathbf{A} - \lambda \mathbf{I}) = 0

\begin{vmatrix} -2 - \lambda & 1 \\ 0 & -2 - \lambda \end{vmatrix} = (-2 - \lambda)^2 = 0

\lambda = -2 is a repeated eigenvalue.

To find the eigenvectors, solve (\mathbf{A} + 2\mathbf{I}) \mathbf{v} = \mathbf{0}:

\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \mathbf{v} = \mathbf{0}

This gives v_2 = 0, so an eigenvector corresponding to \lambda = -2 is:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

Since \lambda = -2 is a repeated eigenvalue with only one linearly independent eigenvector, find a generalized eigenvector by solving (\mathbf{A} + 2\mathbf{I}) \mathbf{v} = \mathbf{v}_1:

\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

This gives v_2 = 1. So a generalized eigenvector is:

\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

(d) Determine All Straight-Line Solutions

Straight-line solutions are of the form \mathbf{x}(t) = \mathbf{v} e^{\lambda t}. Since there is only one eigenvector, we have a solution:

\mathbf{x}_1(t) = c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t}

(e) General Solution Using the Generalized Eigenvector

Since there is a repeated eigenvalue and a generalized eigenvector, the general solution takes the form:

\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{-2t} + c_2 (\mathbf{v}_2 t e^{-2t} + \mathbf{v}_1 e^{-2t})

Substituting \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}:

\mathbf{x}(t) = c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t} + c_2 \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} t e^{-2t} + \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t} \right)

Simplifying, the general solution is:

\mathbf{x}(t) = \left( c_1 + c_2 \right) \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t} + c_2 \begin{bmatrix} 0 \\ t \end{bmatrix} e^{-2t}

(f) Solve the Initial-Value Problem

Given \mathbf{x}(0) = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, apply the initial conditions:

\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \left( c_1 + c_2 \right) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix}

This gives the system:

c_1 + c_2 = 1

c_2 = 1

Solving these, c_2 = 1 and c_1 = 0. So, the solution to the initial-value problem is:

\mathbf{x}(t) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t} + \begin{bmatrix} 0 \\ t \end{bmatrix} e^{-2t} = \begin{bmatrix} e^{-2t} \\ t e^{-2t} \end{bmatrix}

Interpretation of the Solution

The solution \mathbf{x}(t) = \begin{bmatrix} e^{-2t} \\ t e^{-2t} \end{bmatrix} shows that as t \to \infty, both components approach zero because of the e^{-2t} factor. However, the term t e^{-2t} causes the solution to decay slower in the x_2-direction, resulting in a solution that initially grows in the x_2-direction before eventually decaying to zero.

Final Answer:

(a) An explicit matrix \mathbf{A} is

\mathbf{A} = \begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}

(b) The only constant solution is

\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

(c) The eigenvalue is \lambda = -2 (repeated); corresponding eigenvectors are

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

with generalized eigenvector

\mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

(d) A straight-line solution is

\mathbf{x}_1(t) = c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t}

(e) The general solution is

\mathbf{x}(t) = \left( c_1 + c_2 \right) \begin{bmatrix} 1 \\ 0 \end{bmatrix} e^{-2t} + c_2 \begin{bmatrix} 0 \\ t \end{bmatrix} e^{-2t}

(f) The solution to the initial-value problem is

\mathbf{x}(t) = \begin{bmatrix} e^{-2t} \\ t e^{-2t} \end{bmatrix}

Interpretation: The term t e^{-2t} results in an initial growth in the x_2-direction before the solution eventually decays to zero, indicating a slower decay in the x_2-component due to the generalized eigenvector.

3.2.17

Problem:

Convert the second-order differential equation

y'' + y' - 12y = 0

into a system of first-order equations through an appropriate substitution.

Solution:

To convert this into a first-order system, let:

x_1 = y

x_2 = y'

x_1' = y' = x_2

x_2' = y'' = -y' + 12y = -x_2 + 12x_1

The system can be written as:

x_1' = x_2

x_2' = -x_2 + 12x_1

or in matrix form:

\mathbf{x}' = \begin{bmatrix} 0 & 1 \\ 12 & -1 \end{bmatrix} \mathbf{x}

where

\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} y \\ y' \end{bmatrix}

Final Answer:

The equivalent system of first-order equations is:

\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}' = \begin{bmatrix} 0 & 1 \\ 12 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}

SECTION 3.3

Assigned: 4, 5, 11, 12

3.3.4

Problem:

Use the Wronskian to determine if the given set of vector functions is linearly independent:

\mathbf{x}_1(t) = \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix}, \quad \mathbf{x}_2(t) = \begin{bmatrix} e^{2t} \\ 2e^{2t} \end{bmatrix}

Solution:

To determine if \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent, calculate the Wronskian.

Set Up the Wronskian

The Wronskian W(\mathbf{x}_1, \mathbf{x}_2) of two vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) is defined as:

W(\mathbf{x}_1, \mathbf{x}_2) = \det \begin{bmatrix} e^{-t} & e^{2t} \\ -e^{-t} & 2e^{2t} \end{bmatrix}

Calculate the Determinant

Expanding the determinant:

W(\mathbf{x}_1, \mathbf{x}_2) = \left( e^{-t} \cdot 2e^{2t} \right) - \left( e^{2t} \cdot -e^{-t} \right)

= 2e^{t} + e^{t} = 3e^{t}

Result

Since W(\mathbf{x}_1, \mathbf{x}_2) = 3e^t, which is nonzero for all t, the vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent.

Final Answer:

The vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent because the Wronskian W(\mathbf{x}_1, \mathbf{x}_2) = 3e^t is nonzero for all t.

3.3.5

Problem:

Use the Wronskian to determine if the given set of vector functions is linearly independent:

\mathbf{x}_1(t) = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix}, \quad \mathbf{x}_2(t) = \begin{bmatrix} \sin t \\ \cos t \end{bmatrix}

Solution:

To determine if \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent, calculate the Wronskian.

Set Up the Wronskian

The Wronskian W(\mathbf{x}_1, \mathbf{x}_2) of two vector functions \mathbf{x}_1(t) and \mathbf{x}_2(t) is defined as:

W(\mathbf{x}_1, \mathbf{x}_2) = \det \begin{bmatrix} \cos t & \sin t \\ \sin t & \cos t \end{bmatrix}

Calculate the Determinant

Expanding the determinant:

W(\mathbf{x}_1, \mathbf{x}_2) = \left( \cos t \cdot \cos t \right) - \left( \sin t \cdot \sin t \right)

= \cos^2 t - \sin^2 t

Using the trigonometric identity \cos^2 t - \sin^2 t = \cos(2t):

W(\mathbf{x}_1, \mathbf{x}_2) = \cos(2t)

Theorem 3.3.5

According to Theorem 3.3.5, a set of functions is linearly independent on an interval I if the Wronskian is nonzero at at least one point within I.

In this case, W(\mathbf{x}_1, \mathbf{x}_2) = \cos(2t), which is not identically zero on any open interval. Although \cos(2t) becomes zero at specific points (e.g., t = \frac{\pi}{4}, \frac{3\pi}{4}, \ldots), it is nonzero for many other points within any interval.

Final Answer:

Final Answer:

The vector functions \mathbf{x}_1(t) = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix} and \mathbf{x}_2(t) = \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} are linearly independent.

3.3.11

Problem:

Given the system \mathbf{x}' = A \mathbf{x} with matrix

A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}

1. Compute the eigenvalues and eigenvectors of A. Explain why you have found one linearly independent solution to the system, but still need to determine another.

2. Verify through direct substitution that \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} is a solution to the given system \mathbf{x}' = A \mathbf{x}.

3. Show that the solution you found in (a) and the solution \mathbf{x}_2(t) in (b) are linearly independent, and hence state the general solution to the system.

4. Solve the initial value problem (IVP) with the initial condition \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}.

Solution:

(a) Compute the Eigenvalues and Eigenvectors, and Explain the Need for Another Solution

1. Find the Eigenvalues:

   To find the eigenvalues of A, solve \det(A - \lambda I) = 0:

   A - \lambda I = \begin{bmatrix} 3 - \lambda & 1 \\ 0 & 3 - \lambda \end{bmatrix}

   \det(A - \lambda I) = (3 - \lambda)(3 - \lambda) - (1)(0) = (3 - \lambda)^2 = 0

   This gives a repeated eigenvalue \lambda = 3 with algebraic multiplicity 2.

2. Find the Eigenvectors:

   To find the eigenvectors corresponding to \lambda = 3, solve (A - 3I) \mathbf{v} = 0:

   A - 3I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}

   Solving \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, we find that any vector of the form \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} is an eigenvector. So there is only one linearly independent eigenvector \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.

3. Explain the Need for Another Solution:

   According to Theorem 3.3.3, for an n \times n matrix A, we need n linearly independent eigenvectors to form a complete set of solutions for the system \mathbf{x}' = A \mathbf{x}. Each eigenvector \mathbf{v}_i with its corresponding eigenvalue \lambda_i provides a term c_i e^{\lambda_i t} \mathbf{v}_i in the general solution.

   In our problem, the matrix A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} has a repeated eigenvalue \lambda = 3 but only one eigenvector. This is insufficient to meet the requirement of Theorem 3.3.3 for two linearly independent solutions (since A is 2 \times 2). Therefore, we need to find a second solution using a generalized eigenvector to form a complete, linearly independent solution set.

(b) Verify \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} as a Solution

To verify that \mathbf{x}_2(t) is a solution, compute \mathbf{x}_2'(t) and check if \mathbf{x}_2'(t) = A \mathbf{x}_2(t).

1. Calculate \mathbf{x}_2'(t):

   \mathbf{x}_2(t) = \begin{bmatrix} t e^{3t} \\ e^{3t} \end{bmatrix}

   Differentiating each component with respect to t:

   \mathbf{x}_2'(t) = \begin{bmatrix} \frac{d}{dt} (t e^{3t}) \\ \frac{d}{dt} (e^{3t}) \end{bmatrix} = \begin{bmatrix} e^{3t} + 3t e^{3t} \\ 3e^{3t} \end{bmatrix}

2. Compute A \mathbf{x}_2(t): A \mathbf{x}_2(t) = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} t e^{3t} \\ e^{3t} \end{bmatrix} = \begin{bmatrix} 3(t e^{3t}) + e^{3t} \\ 3e^{3t} \end{bmatrix} = \begin{bmatrix} e^{3t} + 3t e^{3t} \\ 3e^{3t} \end{bmatrix}

Since \mathbf{x}_2'(t) = A \mathbf{x}_2(t), \mathbf{x}_2(t) is indeed a solution.

(c) Show Linear Independence and State the General Solution

The solutions are:

\mathbf{x}_1(t) = e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix}

To show that \mathbf{x}_1(t) and \mathbf{x}_2(t) are linearly independent, consider a linear combination:

c_1 \mathbf{x}_1(t) + c_2 \mathbf{x}_2(t) = 0

Since \mathbf{x}_1(t) and \mathbf{x}_2(t) are not multiples of each other, they form a linearly independent set (you can verify this by calculating the Wronskian). So the general solution to the system is:

\mathbf{x}(t) = c_1 e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \left( t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right)

(d) Solve the IVP with Initial Condition \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

Using the initial condition \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}, substitute t = 0 into the general solution:

\mathbf{x}(0) = c_1 e^{0} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \left( 0 \cdot e^{0} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{0} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right)

= c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}

From this, c_1 = 3 and c_2 = 2.

Final Answer:

(a) The matrix A has a repeated eigenvalue \lambda = 3 with only one corresponding eigenvector:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

According to Theorem 3.3.3, to form a complete solution for the system \mathbf{x}' = A \mathbf{x}, we need two linearly independent solutions (since A is a 2 \times 2 matrix). However, since we only have one eigenvector, we must find a second solution by using a generalized eigenvector.

(b) By direct substitution

\mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix}

is a solution to the system \mathbf{x}' = A \mathbf{x}.

(c) The solutions \mathbf{x}_1(t) = e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \mathbf{x}_2(t) = t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} are linearly independent. Therefore, the general solution to the system is:

\mathbf{x}(t) = c_1 e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \left( t e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right)

(d) Solving the initial value problem with \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}, we find c_1 = 3 and c_2 = 2. So the solution to the IVP is:

\mathbf{x}(t) = e^{3t} \begin{bmatrix} 3 + 2t \\ 2 \end{bmatrix}

3.3.12

Problem:

Let \mathbf{x}' = A \mathbf{x} be given by the matrix

A = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}

(a) Compute the eigenvalues and eigenvectors of A. Explain why, despite the repeated eigenvalue, two linearly independent solutions to the system are found.

(b) State the general solution to the system.

(c) Solve the initial value problem with the initial condition \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}.

(d) Explain how the original system could be solved without using eigenvalues and eigenvectors.

Solution:

(a) Compute Eigenvalues and Eigenvectors

1. Finding the Eigenvalues:

   To find the eigenvalues of A, solve the characteristic equation \det(A - \lambda I) = 0:

   A - \lambda I = \begin{bmatrix} 3 - \lambda & 0 \\ 0 & 3 - \lambda \end{bmatrix}

   \det(A - \lambda I) = (3 - \lambda)(3 - \lambda) = (3 - \lambda)^2 = 0

   This gives the repeated eigenvalue \lambda = 3.

2. Finding the Eigenvectors:

   Find the eigenvectors corresponding to \lambda = 3 by solving (A - 3I) \mathbf{v} = 0:

   A - 3I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

   This matrix has infinitely many solutions since any vector \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} in \mathbb{R}^2 satisfies the equation. Two linearly independent vectors can be chosen, such as:

   \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

3. Explanation:

   Despite having a repeated eigenvalue, two linearly independent solutions exist because the matrix A scales the standard basis vectors by the same factor (3). This means that any vector in \mathbb{R}^2 can be represented using the chosen eigenvectors.

(b) State the General Solution to the System

Given the two linearly independent eigenvectors, the general solution is:

\mathbf{x}(t) = c_1 e^{3t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 0 \\ 1 \end{bmatrix}

Simplifying this gives:

\mathbf{x}(t) = e^{3t} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

(c) Solve the IVP with Initial Condition

To solve the initial value problem with \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix}, substitute t = 0 into the general solution:

\mathbf{x}(0) = e^{0} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

This gives c_1 = 3 and c_2 = 2. So the solution to the IVP is:

\mathbf{x}(t) = e^{3t} \begin{bmatrix} 3 \\ 2 \end{bmatrix}

(d) Solve the Original System Without Using Eigenvalues and Eigenvectors

Since A is a diagonal matrix with constant entries, the system can be separated into two independent differential equations:

1. For the first component x_1:

   x_1' = 3x_1

   This is a first-order linear differential equation, with solution:

   x_1(t) = c_1 e^{3t}

2. For the second component x_2: x_2' = 3x_2 Similarly, this gives: x_2(t) = c_2 e^{3t}

Putting both components together yields:

\mathbf{x}(t) = \begin{bmatrix} c_1 e^{3t} \\ c_2 e^{3t} \end{bmatrix} = e^{3t} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

This method solves the system directly by leveraging the diagonal structure of A, without requiring eigenvalues or eigenvectors.

Final Answer:

(a) The matrix A has an eigenvalue \lambda = 3, with two linearly independent eigenvectors \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. The two solutions exist because A scales both basis vectors by the same factor, 3.

(b) The general solution to the system is:

\mathbf{x}(t) = e^{3t} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

(c) The solution to the initial value problem with \mathbf{x}(0) = \begin{bmatrix} 3 \\ 2 \end{bmatrix} is:

\mathbf{x}(t) = e^{3t} \begin{bmatrix} 3 \\ 2 \end{bmatrix}

(d) Without using eigenvalues, the system can be solved by treating x_1 and x_2 as separate differential equations, leading to the same general solution:

\mathbf{x}(t) = e^{3t} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

SECTION 3.4

Assigned: 1, 3, 7, 13

3.4.1

Problem:

Consider the system of differential equations \mathbf{x}' = A \mathbf{x} given by:

A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}

1. (a) Determine the general solution to the system \mathbf{x}' = A \mathbf{x}.
2. (b) Classify the stability of all equilibrium solutions to the system.
3. (c) Sketch all straight-line solutions to the system and plot several nonlinear trajectories in the phase plane.

Solution:

Part (a): Determine the General Solution

Find the Eigenvalues of A

The eigenvalues \lambda are obtained by solving the characteristic equation:

\det(A - \lambda I) = 0

Calculate A - \lambda I:

A - \lambda I = \begin{bmatrix} 2 - \lambda & -1 \\ 3 & -2 - \lambda \end{bmatrix}

eterminant:

\det(A - \lambda I) = (2 - \lambda)(-2 - \lambda) - (-1)(3)

Expanding the expression:

= \lambda^2 - 1

Setting the determinant equal to zero:

\lambda^2 - 1 = 0

Solving for \lambda:

\lambda^2 = 1 \quad \Rightarrow \quad \lambda = 1 \quad \text{or} \quad \lambda = -1

Find the Eigenvectors Corresponding to Each Eigenvalue

For \lambda = 1:

Solve (A - \lambda I) \mathbf{v} = \mathbf{0}:

A - \lambda I = \begin{bmatrix} 1 & -1 \\ 3 & -3 \end{bmatrix}

Set up the system:

\begin{cases} 1 \cdot v_1 - 1 \cdot v_2 = 0 \\ 3 \cdot v_1 - 3 \cdot v_2 = 0 \end{cases}

This reduces to v_1 = v_2. Choosing v_1 = 1, then v_2 = 1:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

For \lambda = -1:

Solve (A + I) \mathbf{v} = \mathbf{0}:

A + I = \begin{bmatrix} 3 & -1 \\ 3 & -1 \end{bmatrix}

Set up the system:

\begin{cases} 3 v_1 - v_2 = 0 \end{cases}

This reduces to v_2 = 3 v_1. Choosing v_1 = 1, then v_2 = 3:

\mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}

General Solution

The general solution is expressed as a linear combination of the eigenvectors scaled by their respective exponential terms:

\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2

Substituting the eigenvalues and eigenvectors:

\mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ 3 \end{bmatrix}

where c_1 and c_2 are arbitrary constants determined by initial conditions.

Part (b): Classify the Stability of Equilibrium Solutions

The equilibrium solution of the system \mathbf{x}' = A \mathbf{x} is at \mathbf{x} = \mathbf{0}.

Stability Analysis:

• Eigenvalues:
  • \lambda_1 = 1 (positive real part)
  • \lambda_2 = -1 (negative real part)

Since one eigenvalue is positive and the other is negative, the equilibrium at the origin is classified as a saddle point.

Conclusion:
The equilibrium solution at \mathbf{x} = \mathbf{0} is unstable due to the presence of a positive eigenvalue.

Part (c): Sketch Straight-Line Solutions and Phase Plane Trajectories

Straight-Line Solutions:

Straight-line solutions correspond to trajectories along the eigenvectors.

• Along \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} (Unstable Eigenvector):

  • Trajectories move away from the origin as t \to \infty because e^{t} grows exponentially.
  • Equation of the line: x_2 = x_1.

• Along \mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix} (Stable Eigenvector):

  • Trajectories approach the origin as t \to \infty because e^{-t} decays exponentially.
  • Equation of the line: x_2 = 3 x_1.

Graph

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

# Define the system of ODEs
def dx_dt(X, t):
    x1, x2 = X
    dx1 = 2 * x1 - x2
    dx2 = 3 * x1 - 2 * x2
    return [dx1, dx2]

# Generate grid for direction field
x1_vals = np.linspace(-5, 5, 35)
x2_vals = np.linspace(-5, 5, 35)
X1, X2 = np.meshgrid(x1_vals, x2_vals)

# Calculate slopes for direction field
U = 2 * X1 - X2
V = 3 * X1 - 2 * X2

# Normalize the direction field arrows
speed = np.sqrt(U**2 + V**2)
epsilon = 1e-10
U_norm = U / (speed + epsilon)
V_norm = V / (speed + epsilon)

# Plot the direction field
plt.figure(figsize=(8, 8))
plt.quiver(X1, X2, U_norm, V_norm, angles="xy", color="black", scale=40, alpha=0.4, width=0.0035)

# Plot the straight-line solutions (eigenvectors)
x = np.linspace(-5, 5, 200)
plt.plot(x, x, 'r', linewidth=3, label=r'Unstable, $\lambda=1$')
plt.plot(x, 3 * x, 'b', linewidth=3, label=r'Stable, $\lambda=-1$')

# Plot trajectories for different initial conditions
initial_conditions = [
    [2, 0],
    [-2, -4],
    [-2, 0],
    [2, 4],
]

t_forward = np.linspace(0, 5, 500)
t_backward = np.linspace(0, -5, 500)

colors = ['green', 'purple', 'yellow', 'orange']
for idx, ic in enumerate(initial_conditions):
    # Solve ODE forward in time
    sol_forward = odeint(dx_dt, ic, t_forward)
    plt.plot(sol_forward[:, 0], sol_forward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

    # Solve ODE backward in time
    sol_backward = odeint(dx_dt, ic, t_backward)
    plt.plot(sol_backward[:, 0], sol_backward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

# Set plot limits and labels
plt.xlim(-5, 5)
plt.ylim(-5, 5)
plt.xlabel("$x_1$", fontsize=14)
plt.ylabel("$x_2$", fontsize=14)
plt.title(r"Direction Field and Trajectories for $\mathbf{x'} = A\mathbf{x}$", fontsize=18)

# Add grid and legend
plt.grid(True, linestyle='--', alpha=0.5)
plt.legend(loc="best", fontsize=12, framealpha=1.0, frameon=True)

# Display the plot
plt.tight_layout()
plt.show()

Final Answer:

(a) The general solution to the system is:

\mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ 3 \end{bmatrix}

where c_1 and c_2 are constants determined by initial conditions.

(b) The equilibrium solution at \mathbf{x} = \mathbf{0} is a saddle point and is unstable due to one positive and one negative eigenvalue.

(c) The straight-line solutions are along the lines x_2 = x_1 (unstable manifold) and x_2 = 3 x_1 (stable manifold). In the phase plane, trajectories exhibit saddle point behavior, moving away from the origin along the unstable direction and curving towards the stable direction when traced backward in time.

3.4.3

Problem:

Consider the system of differential equations \mathbf{x}' = A \mathbf{x} given by:

A = \begin{bmatrix} -3 & 2 \\ 2 & -3 \end{bmatrix}

1. (a) Determine the general solution to the system \mathbf{x}' = A \mathbf{x}.
2. (b) Classify the stability of all equilibrium solutions to the system.
3. (c) Sketch all straight-line solutions to the system and plot several nonlinear trajectories in the phase plane.

Solution:

Part (a): Determine the General Solution

Find the Eigenvalues of A

The eigenvalues \lambda are obtained by solving the characteristic equation:

\det(A - \lambda I) = 0

Calculate A - \lambda I:

A - \lambda I = \begin{bmatrix} -3 - \lambda & 2 \\ 2 & -3 - \lambda \end{bmatrix}

Determinant:

\det(A - \lambda I) = (\lambda + 3)^2 - 4 = \lambda^2 + 6\lambda + 5

Setting the determinant equal to zero:

\lambda^2 + 6\lambda + 5 = 0

Solving for \lambda:

\lambda = -1 \quad \text{and} \quad \lambda = -5

Find the Eigenvectors Corresponding to Each Eigenvalue

For \lambda = -1:

Solve (A + I) \mathbf{v} = \mathbf{0}:

A + I = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}

Set up the system:

\begin{cases} -2 v_1 + 2 v_2 = 0 \\ 2 v_1 - 2 v_2 = 0 \end{cases}

This reduces to v_1 = v_2. Choosing v_1 = 1, then v_2 = 1:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

For \lambda = -5:

Solve (A + 5I) \mathbf{v} = \mathbf{0}:

A + 5I = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}

Set up the system:

\begin{cases} 2 v_1 + 2 v_2 = 0 \end{cases}

This reduces to v_1 = -v_2. Choosing v_1 = 1, then v_2 = -1:

\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}

General Solution

The general solution is expressed as a linear combination of the eigenvectors scaled by their respective exponential terms:

\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2

Substituting the eigenvalues and eigenvectors:

\mathbf{x}(t) = c_1 e^{-t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-5t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

where c_1 and c_2 are arbitrary constants determined by initial conditions.

Part (b): Classify the Stability of Equilibrium Solutions

The equilibrium solution of the system \mathbf{x}' = A \mathbf{x} is at \mathbf{x} = \mathbf{0}.

Stability Analysis:

• Eigenvalues:
  • \lambda_1 = -1
  • \lambda_2 = -5

Since both eigenvalues are negative, the equilibrium at the origin is classified as a stable attracting node.

Conclusion:
The equilibrium solution at \mathbf{x} = \mathbf{0} is stable because all eigenvalues have negative real parts.

Part (c): Sketch Straight-Line Solutions and Phase Plane Trajectories

Straight-Line Solutions:

Straight-line solutions correspond to trajectories along the eigenvectors.

• Along \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}:

  • Trajectories approach the origin as t \to \infty because e^{-t} decays exponentially.
  • Equation of the line: x_2 = x_1.

• Along \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}:

  • Trajectories approach the origin as t \to \infty because e^{-5t} decays even faster.
  • Equation of the line: x_2 = -x_1.

Graph

# Define the system of ODEs
def dx_dt(X, t):
    x1, x2 = X
    dx1 = -3 * x1 + 2 * x2
    dx2 = 2 * x1 - 3 * x2
    return [dx1, dx2]

# Generate grid for direction field
x1_vals = np.linspace(-5, 5, 35)
x2_vals = np.linspace(-5, 5, 35)
X1, X2 = np.meshgrid(x1_vals, x2_vals)

# Calculate slopes for direction field
U = -3 * X1 + 2 * X2
V = 2 * X1 - 3 * X2

# Normalize the direction field arrows
speed = np.sqrt(U**2 + V**2)
epsilon = 1e-10  # Small value to prevent division by zero
U_norm = U / (speed + epsilon)
V_norm = V / (speed + epsilon)

# Plot the direction field
plt.figure(figsize=(8, 8))
plt.quiver(X1, X2, U_norm, V_norm, angles="xy", color="black", scale=40, alpha=0.4, width=0.0035)

# Plot the straight-line solutions (eigenvectors)
x = np.linspace(-5, 5, 200)
plt.plot(x, x, 'r', linewidth=3, label=r'Stable direction, $\lambda=-1$')
plt.plot(x, -x, 'b', linewidth=3, label=r'Stable direction, $\lambda=-5$')

# Plot trajectories for different initial conditions
initial_conditions = [
    [2, 0],
    [-2, -4],
    [-2, 0],
    [2, 4],
]

t_forward = np.linspace(0, 5, 500)
t_backward = np.linspace(0, -5, 500)

colors = ['green', 'purple', 'yellow', 'orange']
for idx, ic in enumerate(initial_conditions):
    # Solve ODE forward in time
    sol_forward = odeint(dx_dt, ic, t_forward)
    plt.plot(sol_forward[:, 0], sol_forward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

    # Solve ODE backward in time
    sol_backward = odeint(dx_dt, ic, t_backward)
    plt.plot(sol_backward[:, 0], sol_backward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

# Set plot limits and labels
plt.xlim(-5, 5)
plt.ylim(-5, 5)
plt.xlabel("$x_1$", fontsize=14)
plt.ylabel("$x_2$", fontsize=14)
plt.title(r"Direction Field and Trajectories for $\mathbf{x'} = A\mathbf{x}$", fontsize=18)

# Add grid and legend
plt.grid(True, linestyle='--', alpha=0.5)
plt.legend(loc="best", fontsize=12, framealpha=1.0, frameon=True)

# Display the plot
plt.tight_layout()
plt.show()

Final Answer:

(a) The general solution to the system is:

\mathbf{x}(t) = c_1 e^{-t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-5t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

where c_1 and c_2 are constants determined by initial conditions.

(b) The equilibrium solution at \mathbf{x} = \mathbf{0} is a stable attracting node because both eigenvalues are negative.

(c) The straight-line solutions are along the lines x_2 = x_1 and x_2 = -x_1. In the phase plane, trajectories exhibit stable attracting node behavior,

with all paths approaching the origin.

3.4.7

Problem:

Consider the system of differential equations \mathbf{x}' = A \mathbf{x} with the general solution:

\mathbf{x}(t) = c_1 e^{4t} \begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_2 e^{-3t} \begin{bmatrix} 1 \\ 2 \end{bmatrix}

1. (a) State the straight-line solutions to the system.
2. (b) Classify the stability of the equilibrium solution at the origin.
3. (c) Sketch all straight-line solutions to the system and plot several trajectories in the phase plane.

Solution:

Part (a): State the Straight-Line Solutions

The given general solution consists of two terms:

1. c_1 e^{4t} \begin{bmatrix} -1 \\ 2 \end{bmatrix}
2. c_2 e^{-3t} \begin{bmatrix} 1 \\ 2 \end{bmatrix}

Each term represents a straight-line solution along the direction of its corresponding eigenvector:

• Unstable direction:

  • Eigenvalue \lambda = 4 with eigenvector \begin{bmatrix} -1 \\ 2 \end{bmatrix}.
  • Straight-line solution: x_2 = -2x_1.
  • Since \lambda = 4 > 0, trajectories move away from the origin along this line as t \to \infty.

• Stable direction:

  • Eigenvalue \lambda = -3 with eigenvector \begin{bmatrix} 1 \\ 2 \end{bmatrix}.
  • Straight-line solution: x_2 = 2x_1.
  • Since \lambda = -3 < 0, trajectories move toward the origin along this line as t \to \infty.

Part (b): Classify the Stability of the Equilibrium Solution

The equilibrium solution of the system \mathbf{x}' = A \mathbf{x} is at \mathbf{x} = \mathbf{0}.

Stability Analysis:

• Eigenvalues:
  • \lambda_1 = 4
  • \lambda_2 = -3

Since one eigenvalue is positive and the other is negative, the equilibrium at the origin is classified as a saddle point.

Conclusion:
The equilibrium solution at \mathbf{x} = \mathbf{0} is unstable due to the presence of a positive eigenvalue.

Part (c): Sketch Straight-Line Solutions and Phase Plane Trajectories

Straight-Line Solutions:

Straight-line solutions correspond to trajectories along the eigenvectors.

• Along \mathbf{v}_1 = \begin{bmatrix} -1 \\ 2 \end{bmatrix} (Unstable Eigenvector):

  • Trajectories move away from the origin as t \to \infty because e^{4t} grows exponentially.
  • Equation of the line: x_2 = -2x_1.

• Along \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} (Stable Eigenvector):

  • Trajectories approach the origin as t \to \infty because e^{-3t} decays exponentially.
  • Equation of the line: x_2 = 2x_1.

Graph

# Corrected system of ODEs using the matrix A
def dx_dt(X, t):
    x1, x2 = X
    dx1 = 0.5 * x1 - 1.75 * x2
    dx2 = -7 * x1 + 0.5 * x2
    return [dx1, dx2]

# Generate grid for direction field
x1_vals = np.linspace(-10, 10, 40)
x2_vals = np.linspace(-10, 10, 40)
X1, X2 = np.meshgrid(x1_vals, x2_vals)

# Calculate slopes for direction field
U = 0.5 * X1 - 1.75 * X2
V = -7 * X1 + 0.5 * X2

# Normalize the direction field arrows
speed = np.sqrt(U**2 + V**2)
epsilon = 1e-10  # Small value to prevent division by zero
U_norm = U / (speed + epsilon)
V_norm = V / (speed + epsilon)

# Plot the direction field
plt.figure(figsize=(8, 8))
plt.quiver(X1, X2, U_norm, V_norm, angles="xy", color="gray", scale=40, alpha=0.4, width=0.0035)

# Plot the straight-line solutions (eigenvectors)
x = np.linspace(-10, 10, 400)
plt.plot(x, -2 * x, 'r', linewidth=3, label=r'Unstable direction, $\lambda=4$')
plt.plot(x, 2 * x, 'b', linewidth=3, label=r'Stable direction, $\lambda=-3$')

# Plot trajectories for different initial conditions
initial_conditions = [
    [2, 0],
    [-2, 0],
    [0, 5],
    [0, -5],
]

t_forward = np.linspace(0, 2, 500)
t_backward = np.linspace(0, -2, 500)

colors = ['green', 'purple', 'orange', 'cyan']
for idx, ic in enumerate(initial_conditions):
    # Solve ODE forward in time
    sol_forward = odeint(dx_dt, ic, t_forward)
    plt.plot(sol_forward[:, 0], sol_forward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

    # Solve ODE backward in time
    sol_backward = odeint(dx_dt, ic, t_backward)
    plt.plot(sol_backward[:, 0], sol_backward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

# Set plot limits and labels
plt.xlim(-10, 10)
plt.ylim(-10, 10)
plt.xlabel("$x_1$", fontsize=14)
plt.ylabel("$x_2$", fontsize=14)
plt.title(r"Direction Field and Trajectories for $\mathbf{x}' = A\mathbf{x}$", fontsize=16)

# Add grid and legend
plt.grid(True, linestyle='--', alpha=0.5)
plt.legend(loc="upper left", fontsize=12, framealpha=1.0, frameon=True)

# Display the plot
plt.tight_layout()
plt.show()

Final Answer:

(a) The straight-line solutions to the system are along the lines:

• x_2 = -2x_1 (Unstable direction, \lambda = 4)
• x_2 = 2x_1 (Stable direction, \lambda = -3)

(b) The equilibrium solution at \mathbf{x} = \mathbf{0} is classified as a saddle point and is unstable because one eigenvalue is positive.

(c) In the phase plane, trajectories exhibit saddle point behavior. Trajectories approach the origin along the line x_2 = 2x_1 (stable direction) and move away from the origin along x_2 = -2x_1 (unstable direction).

3.4.13

Problem:

Solve the initial value problem (IVP) for \mathbf{x}' = A \mathbf{x} with the given matrix A and initial condition.

Given:

A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}, \quad \mathbf{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}

Solution:

Determine the General Solution

Find the Eigenvalues of A

The eigenvalues \lambda are obtained by solving the characteristic equation:

\det(A - \lambda I) = 0

Calculate A - \lambda I:

A - \lambda I = \begin{bmatrix} 2 - \lambda & -1 \\ 3 & -2 - \lambda \end{bmatrix}

Expanding the determinant:

\det(A - \lambda I) = (2 - \lambda)(-2 - \lambda) - (-1)(3)

Simplifying:

= \lambda^2 - 1

Setting the determinant equal to zero:

\lambda^2 - 1 = 0

Solving for \lambda:

\lambda = 1 \quad \text{and} \quad \lambda = -1

Find the Eigenvectors Corresponding to Each Eigenvalue

For \lambda = 1:

Solve (A - \lambda I) \mathbf{v} = \mathbf{0}:

A - \lambda I = \begin{bmatrix} 1 & -1 \\ 3 & -3 \end{bmatrix}

Set up the system:

\begin{cases} 1 \cdot v_1 - 1 \cdot v_2 = 0 \\ 3 \cdot v_1 - 3 \cdot v_2 = 0 \end{cases}

This reduces to v_1 = v_2. Choosing v_1 = 1, then v_2 = 1:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

For \lambda = -1:

Solve (A + I) \mathbf{v} = \mathbf{0}:

A + I = \begin{bmatrix} 3 & -1 \\ 3 & -1 \end{bmatrix}

Set up the system:

\begin{cases} 3 v_1 - v_2 = 0 \end{cases}

This reduces to v_2 = 3 v_1. Choosing v_1 = 1, then v_2 = 3:

\mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}

General Solution

The general solution is expressed as a linear combination of the eigenvectors scaled by their respective exponential terms:

\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2

Substituting the eigenvalues and eigenvectors:

\mathbf{x}(t) = c_1 e^{t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ 3 \end{bmatrix}

where c_1 and c_2 are arbitrary constants determined by initial conditions.

Apply the Initial Condition

Using the initial condition \mathbf{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, substitute t = 0:

\mathbf{x}(0) = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}

This gives the system:

1. c_1 + c_2 = 1
2. c_1 + 3c_2 = 2

Solving these equations, subtract the first equation from the second:

( c_1 + 3c_2) - ( c_1 + c_2) = 2 - 1,

which simplifies to

2c_2 = 1 \Rightarrow c_2 = \frac{1}{2}

Substitute c_2 = \frac{1}{2} into the first equation:

c_1 + \frac{1}{2} = 1 \Rightarrow c_1 = \frac{1}{2}

Final Answer:

The solution to the initial value problem is:

\mathbf{x}(t) = \frac{1}{2} e^t \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \frac{1}{2} e^{-t} \begin{bmatrix} 1 \\ 3 \end{bmatrix}

SECTION 3.5

Assigned: 9, 11, 15, 17

Table 3.1

Eigenvalues Condition	Stability and Classification
0 < \lambda_1 \leq \lambda_2	0 is unstable and called a repelling node
\lambda_1 < 0 < \lambda_2	0 is unstable and called a saddle
\lambda_1 \leq \lambda_2 < 0	0 is stable and called an attracting node
\lambda = a \pm bi and a > 0	0 is unstable and called a spiral source
\lambda = a \pm bi and a = 0	0 is stable and called a center
\lambda = a \pm bi and a < 0	0 is stable and called a spiral sink

3.5.9

Problem:

For the matrix A with characteristic polynomial

p(\lambda) = \lambda^2 + 4,

find the eigenvalues of A and classify the stability of the origin as an equilibrium point of the system given by \mathbf{x}' = A\mathbf{x}.

Solution:

Determine the Eigenvalues

The eigenvalues \lambda are obtained by solving the characteristic equation:

p(\lambda) = \lambda^2 + 4 = 0.

Solving for \lambda:

\lambda^2 = -4

\lambda = \pm 2i.

The eigenvalues are \lambda = 2i and \lambda = -2i, which are purely imaginary (with no real part).

Classify the Stability of the Equilibrium Point

Using Table 3.1, if the eigenvalues are of the form \lambda = a \pm bi with a = 0 (purely imaginary), then the origin is classified as stable and is called a center. This classification indicates that the origin exhibits neutral stability, where solutions oscillate around the origin in a closed, non-decaying pattern.

Final Answer:

The equilibrium point at the origin is stable and is classified as a center, resulting in oscillatory behavior around the origin without moving towards or away from it.

3.5.11

Problem:

For the matrix A with characteristic polynomial

p(\lambda) = \lambda^2 - 10\lambda + 9,

find the eigenvalues of A and classify the stability of the origin as an equilibrium point of the system given by \mathbf{x}' = A\mathbf{x}.

Solution:

Determine the Eigenvalues

The eigenvalues \lambda are obtained by solving the characteristic equation:

p(\lambda) = \lambda^2 - 10\lambda + 9 = 0.

This is a quadratic equation, which can be factored as:

(\lambda - 9)(\lambda - 1) = 0.

Solving for \lambda:

\lambda = 9 \quad \text{and} \quad \lambda = 1.

The eigenvalues are \lambda = 9 and \lambda = 1, both of which are positive.

Classify the Stability of the Equilibrium Point

Using Table 3.1, when both eigenvalues are positive (0 < \lambda_1 \leq \lambda_2), the origin is classified as unstable and is called a repelling node. This indicates that trajectories move away from the origin as t \to \infty.

Conclusion

The equilibrium point at the origin is unstable and is classified as a repelling node, meaning that solutions will diverge from the origin over time.

Final Answer:

The origin is unstable and is classified as a repelling node.

3.5.15

Problem:

Consider the system of differential equations \mathbf{x}' = A \mathbf{x} with the given matrix

A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}

(a) Determine the general solution to the system \mathbf{x}' = A \mathbf{x}.
(b) Classify the stability of the equilibrium solution at the origin.
(c) Determine how many straight-line solutions this system of equations has, and explain why.
(d) Sketch all straight-line solutions to the system and plot several trajectories in the phase plane.

Solution:

Part (a): Determine the General Solution

Find the Eigenvalues of A

To solve \mathbf{x}' = A \mathbf{x}, start by finding the eigenvalues \lambda of matrix A:

\det(A - \lambda I) = 0

Compute A - \lambda I:

A - \lambda I = \begin{bmatrix} -\lambda & -2 \\ 2 & -\lambda \end{bmatrix}

Calculate the determinant:

\det(A - \lambda I) = (-\lambda)(-\lambda) - (-2)(2) = \lambda^2 + 4 = 0

Solve for \lambda:

\lambda^2 = -4 \implies \lambda = \pm 2i

Eigenvalues: \lambda_1 = 2i, \lambda_2 = -2i.

Find the Eigenvectors of A

For \lambda = 2i:

Solve (A - \lambda I)\mathbf{v} = \mathbf{0}:

\begin{bmatrix} -2i & -2 \\ 2 & -2i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

This equation implies v_2 = -i v_1, so choosing v_1 = 1, we get:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ -i \end{bmatrix}

For \lambda = -2i:

Solve (A + 2i I)\mathbf{v} = \mathbf{0}:

\begin{bmatrix} 2i & -2 \\ 2 & 2i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

This equation implies v_2 = i v_1, so choosing v_1 = 1, we get:

\mathbf{v}_2 = \begin{bmatrix} 1 \\ i \end{bmatrix}

Construct the General Solution

The general solution using complex exponentials is:

\mathbf{x}(t) = c_1 e^{2i t} \begin{bmatrix} 1 \\ -i \end{bmatrix} + c_2 e^{-2i t} \begin{bmatrix} 1 \\ i \end{bmatrix}

Using Euler’s formula e^{i\theta} = \cos \theta + i \sin \theta, rewrite the solution in real terms:

\mathbf{x}(t) = c_1 \left(\cos(2t) + i \sin(2t)\right) \begin{bmatrix} 1 \\ -i \end{bmatrix} + c_2 \left(\cos(2t) - i \sin(2t)\right) \begin{bmatrix} 1 \\ i \end{bmatrix}

Expanding and separating real and imaginary parts, the solution can be expressed as:

\mathbf{x}(t) = c_1 \begin{bmatrix} \cos(2t) \\ \sin(2t) \end{bmatrix} + c_2 \begin{bmatrix} -\sin(2t) \\ \cos(2t) \end{bmatrix}

The real general solution is:

\mathbf{x}(t) = c_1 \begin{bmatrix} \cos(2t) \\ \sin(2t) \end{bmatrix} + c_2 \begin{bmatrix} -\sin(2t) \\ \cos(2t) \end{bmatrix}

where c_1 and c_2 are constants determined by initial conditions.

Part (b): Classify the Stability of the Equilibrium Solution

The equilibrium solution is at \mathbf{x} = \mathbf{0}.

Eigenvalues: \lambda = \pm 2i.

Referring to Table 3.1:

Since a = 0 (purely imaginary eigenvalues), the equilibrium at \mathbf{x} = \mathbf{0} is classified as a center.

Conclusion:

• The equilibrium solution at the origin is stable and classified as a center.

Part (c): Determine the Number of Straight-Line Solutions

There are no non-trivial straight-line solutions because the eigenvalues are imaginary, leading to oscillatory solutions rather than motion along straight lines.

Part (d): Graph

Graph

# Define the system of ODEs using the matrix A
def dx_dt(X, t):
    x1, x2 = X
    dx1 = -2 * x2
    dx2 = 2 * x1
    return [dx1, dx2]

# Generate grid for direction field
x1_vals = np.linspace(-10, 10, 40)
x2_vals = np.linspace(-10, 10, 40)
X1, X2 = np.meshgrid(x1_vals, x2_vals)

# Calculate slopes for direction field
U = -2 * X2
V = 2 * X1

# Normalize the direction field arrows
speed = np.sqrt(U**2 + V**2)
epsilon = 1e-10  # Small value to prevent division by zero
U_norm = U / (speed + epsilon)
V_norm = V / (speed + epsilon)

# Plot the direction field
plt.figure(figsize=(8, 8))
plt.quiver(X1, X2, U_norm, V_norm, angles="xy", color="gray", scale=40, alpha=0.4, width=0.0035)

# Plot the circular trajectories for different initial conditions
initial_conditions = [
    [1, 0],
    [5, 0],
    [9, 0],
]

t_forward = np.linspace(0, 10, 500)
t_backward = np.linspace(0, -10, 500)

colors = ['purple', 'cyan', 'orange']
for idx, ic in enumerate(initial_conditions):
    # Solve ODE forward in time
    sol_forward = odeint(dx_dt, ic, t_forward)
    plt.plot(sol_forward[:, 0], sol_forward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

    # Solve ODE backward in time
    sol_backward = odeint(dx_dt, ic, t_backward)
    plt.plot(sol_backward[:, 0], sol_backward[:, 1], color=colors[idx], linestyle='-', linewidth=2)

# Set plot limits and labels
plt.xlim(-10, 10)
plt.ylim(-10, 10)
plt.xlabel("$x_1$", fontsize=14)
plt.ylabel("$x_2$", fontsize=14)
plt.title(r"Direction Field and Trajectories for $\mathbf{x}' = A\mathbf{x}$", fontsize=16)

# Add grid and display the plot
plt.grid(True, linestyle='--', alpha=0.5)
plt.tight_layout()
plt.show()

Final Answer:

(a) The general solution to the system is:

\mathbf{x}(t) = c_1 \begin{bmatrix} \cos(2t) \\ \sin(2t) \end{bmatrix} + c_2 \begin{bmatrix} -\sin(2t) \\ \cos(2t) \end{bmatrix}

(b) The equilibrium solution at the origin is stable and classified as a center.

(c) The system has no non-trivial straight-line solutions because the eigenvalues are purely imaginary, resulting in oscillatory behavior rather than straight-line motion.

(d) The phase plane consists of circular trajectories centered at the origin, with counterclockwise motion.

3.5.17

Problem:

Consider the system of differential equations \mathbf{x}' = A \mathbf{x} with the given matrix

A = \begin{bmatrix} -2 & 1 \\ 0 & -2 \end{bmatrix}

(a) Determine the general solution to the system \mathbf{x}' = A \mathbf{x}.
(b) Classify the stability of the equilibrium solution at the origin.
(c) Determine how many straight-line solutions this system of equations has, and explain why.
(d) Sketch all straight-line solutions to the system and plot several trajectories in the phase plane.

Solution:

Part (a): Determine the General Solution

Find the Eigenvalues of A:

To solve the system \mathbf{x}' = A \mathbf{x}, start by finding the eigenvalues \lambda of matrix A:

\det(A - \lambda I) = 0

Compute A - \lambda I:

A - \lambda I = \begin{bmatrix} -2 - \lambda & 1 \\ 0 & -2 - \lambda \end{bmatrix}

Calculate the determinant:

\det(A - \lambda I) = (-2 - \lambda)(-2 - \lambda) - (1)(0) = (-2 - \lambda)^2

Set the determinant to zero:

(-2 - \lambda)^2 = 0 \implies -2 - \lambda = 0 \implies \lambda = -2

Eigenvalue: \lambda = -2 (repeated eigenvalue of multiplicity 2).

Find the Eigenvector of A:

Solve (A - \lambda I)\mathbf{v} = \mathbf{0}:

(A + 2I)\mathbf{v} = \mathbf{0}

Compute A + 2I:

A + 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}

Set up the equations:

\begin{cases} 0 \cdot v_1 + 1 \cdot v_2 = 0 \implies v_2 = 0 \\ 0 \cdot v_1 + 0 \cdot v_2 = 0 \end{cases}

Thus, the eigenvector corresponding to \lambda = -2 is:

\mathbf{v} = \begin{bmatrix} v_1 \\ 0 \end{bmatrix} = v_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix}

Apply Theorem 3.5.1:

According to Theorem 3.5.1, if A has a repeated eigenvalue \lambda and only one linearly independent eigenvector \mathbf{v}, the general solution to \mathbf{x}' = A \mathbf{x} is:

\mathbf{x}(t) = c_1 e^{\lambda t} \mathbf{v} + c_2 e^{\lambda t} (t \mathbf{v} + \mathbf{u})

where \mathbf{u} satisfies:

(A - \lambda I) \mathbf{u} = \mathbf{v}

Find the Generalized Eigenvector \mathbf{u}:

Solve (A - \lambda I) \mathbf{u} = \mathbf{v}:

Since A - \lambda I = A + 2I:

(A + 2I) \mathbf{u} = \mathbf{v}

Compute A + 2I as before:

A + 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}

Set up the equation:

\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

From the first row:

0 \cdot u_1 + 1 \cdot u_2 = 1 \implies u_2 = 1

The second row gives:

0 \cdot u_1 + 0 \cdot u_2 = 0 \implies 0 = 0

Choose u_1 = 0 for simplicity. Thus, the generalized eigenvector is:

\mathbf{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}

Construct the General Solution:

Using Theorem 3.5.1, substitute \lambda = -2, \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, and \mathbf{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}:

\mathbf{x}(t) = e^{-2t} \left( c_1 \mathbf{v} + c_2 (t \mathbf{v} + \mathbf{u}) \right)

Simplify:

\mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{-2t} \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right)

The general solution is:

\mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} t \\ 1 \end{bmatrix}

where c_1 and c_2 are constants determined by initial conditions.

Part (b): Classify the Stability of the Equilibrium Solution

The equilibrium solution is at \mathbf{x} = \mathbf{0}.

Eigenvalues: \lambda = -2 (real, negative, repeated).

Since both eigenvalues have negative real parts, the equilibrium at the origin is:

• Asymptotically stable (solutions approach the origin as t \to \infty).
• Classified as a stable attracting node.

Part (c): Determine the Number of Straight-Line Solutions

Straight-line solutions correspond to eigenvectors.

• There is one eigenvector \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.
• Therefore, there is one straight-line solution along the direction of \mathbf{v}.

Explanation:

• Since the matrix is defective (only one linearly independent eigenvector), there is only one straight-line solution.
• Solutions not along \mathbf{v} will not remain on a straight line but will curve towards the eigenvector direction as t \to \infty.

Part (d): Sketch All Straight-Line Solutions and Plot Several Trajectories

Straight-Line Solution:

• The straight-line solution lies along the x_1-axis.

• It is given by:

  \mathbf{x}(t) = c e^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix}

Phase Plane Trajectories:

• Trajectories approach the origin along curves that become parallel to the x_1-axis as t \to \infty.
• Solutions starting off the x_1-axis will curve towards it over time.

Graphical Representation:

Below is a Python code snippet that generates the direction field and trajectories for the system:

# Define the system of ODEs
def dx_dt(X, t):
    x1, x2 = X
    dx1 = -2 * x1 + x2
    dx2 = -2 * x2
    return [dx1, dx2]

# Generate grid for direction field
x1_vals = np.linspace(-5, 5, 35)
x2_vals = np.linspace(-5, 5, 35)
X1, X2 = np.meshgrid(x1_vals, x2_vals)

# Calculate slopes for direction field
U = -2 * X1 + X2
V = -2 * X2

# Normalize the direction field arrows
speed = np.sqrt(U**2 + V**2)
epsilon = 1e-10  # Small value to prevent division by zero
U_norm = U / (speed + epsilon)
V_norm = V / (speed + epsilon)

# Plot the direction field
plt.figure(figsize=(8, 8))
plt.quiver(X1, X2, U_norm, V_norm, angles="xy", color="black", scale=40, alpha=0.4, width=0.0035)

# Plot the straight-line solution (eigenvector)
x1_line = np.linspace(-5, 5, 200)
x2_line = np.zeros_like(x1_line)
plt.plot(x1_line, x2_line, 'r', linewidth=3, label='Straight-line solution')

# Plot trajectories for different initial conditions
initial_conditions = [
    [2, 2],
    [-2, -2],
    [2, -2],
    [-2, 2],
]

t_forward = np.linspace(0, 5, 500)
t_backward = np.linspace(0, -5, 500)

colors = ['green', 'purple', 'yellow', 'orange']
for idx, ic in enumerate(initial_conditions):
    # Solve ODE forward in time
    sol_forward = odeint(dx_dt, ic, t_forward)
    plt.plot(sol_forward[:, 0], sol_forward[:, 1], color=colors[idx % len(colors)], linestyle='-', linewidth=2)

    # Solve ODE backward in time
    sol_backward = odeint(dx_dt, ic, t_backward)
    plt.plot(sol_backward[:, 0], sol_backward[:, 1], color=colors[idx % len(colors)], linestyle='-', linewidth=2)

# Set plot limits and labels
plt.xlim(-5, 5)
plt.ylim(-5, 5)
plt.xlabel("$x_1$", fontsize=14)
plt.ylabel("$x_2$", fontsize=14)
plt.title(r"Direction Field and Trajectories for $\mathbf{x'} = A\mathbf{x}$", fontsize=18)

# Add grid and legend
plt.grid(True, linestyle='--', alpha=0.5)
plt.legend(loc="best", fontsize=12, framealpha=1.0, frameon=True)

# Display the plot
plt.tight_layout()
plt.show()

Graph Interpretation:

• The straight red dashed line represents the straight-line solution along the x_1-axis.
• Other trajectories curve towards the x_1-axis as t \to \infty.
• All trajectories move towards the origin, confirming the asymptotic stability.

Final Answer:

(a) The general solution to the system is:

\mathbf{x}(t) = e^{-2t} \begin{bmatrix} c_1 + c_2 t \\ c_2 \end{bmatrix}

(b) The equilibrium solution at the origin is asymptotically stable and classified as a stable attracting node.

(c) The system has one straight-line solution along the x_1-axis, corresponding to the eigenvector \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.

(d) The phase plane shows trajectories approaching the origin, with all solutions eventually aligning with the straight-line solution along the x_1-axis. The single straight-line solution is along the x_1-axis, and other trajectories curve towards it over time.